A parallel plate capacitor made of circular plates each of radius 10.0 cm has a capacitance 200 pF. The capacitor is connected to a 200 V ac supply with an angular frequency of 200 rad/s. The amplitude of the current is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8\times10^{-6} \) A
The capacitor offers a capacitive reactance \( X_C = \frac{1}{\omega C} \) to the alternating current, and the current amplitude is \( I_0 = \frac{V_0}{X_C} = \omega CV_0 \). The 200 V is the amplitude of the supply, as the question takes it, so no \( \sqrt{2} \) factor enters.
\( \displaystyle X_C = \frac{1}{200\,\mathrm{rad\,s^{-1}}\times200\times10^{-12}\,\mathrm{F}} = 2.5\times10^{7}\,\Omega \)
\( \displaystyle I_0 = \frac{V_0}{X_C} = \frac{200\,\mathrm{V}}{2.5\times10^{7}\,\Omega} = 8\times10^{-6}\,\mathrm{A} \)
The radius of the plates is not needed, as the capacitance is already given.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – “… 8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s–1. (a) What is the rms value of the conduction current?”